Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 (AC2, 4302)

Source release: Oxford examination archive, member FOI-20210708-1 - Data/4302.pdf. The cover prints formal code 4302, paper label AC2, the title “Algebra, Analysis and Differential equations II”, and Tuesday 19 June in Trinity Term 2012.

Answer availability checked on 28 October 2026: a bounded search of official Oxford sources found no advertisement of complete or selected-part solutions for this exact paper. The general advertisement of even-year solutions dating back to at least 2016 does not establish coverage of 2012.^[ref]

The minimal polynomial $m_T(t)$ is the unique monic polynomial of least positive polynomial degree satisfying $m_T(T)=0$. Such an annihilating polynomial exists without using the Cayley-Hamilton theorem: the vector space $\mathrm{End}(V)$ has dimension of a vector space equal to $n^2$, so $I,T,\dots ,T^{n^2}$ are linearly dependent. Normalize a nonzero annihilating polynomial to be monic and choose one of least degree. Two distinct monic choices of that degree would have an annihilating difference of lower degree, contradicting minimality. A nonzero constant cannot annihilate $T$ because $V\ne 0$.

Choose a basis $B$ and let $A=[T{]}_B$. The characteristic polynomial isIt is independent of the chosen basis. Indeed a change of basis replaces $A$ by a similar matrix $P^{-1}AP$, andby multiplicativity of the determinant. Thus this monic degree-$n$ polynomial is an invariant of the linear transformation itself.

Since $v$ is a cyclic vector and $V\ne 0$, $v\ne 0$. Finite dimension of a vector space forces a first dependence in the sequence $v,Tv,T^{2}v,\dots$. Let $k\ge 1$ be the smallest index with $T^{k}v$ in the linear span of its predecessors. Thenis linearly independent. Its linear span $W$ is invariant under $T$: all but the last basis candidate map to the next, and the last maps to $T^{k}v\in W$. Induction therefore puts every $T^{j}v$ in $W$. Cyclicity gives $W=V$, so $B$ is a basis and $k=n$.

WriteThe matrix of $T$ in this basis is the companion matrixFor $p(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$, the defining relation says $p(T)v=0$. Since $p(T)$ commutes with $T$, it also annihilates every $T^{j}v$ in $B$, hence $p(T)=0$. No nonzero polynomial of degree less than $n$ can annihilate $v$, because that would contradict linear independence of $B$. Thus $m_T=p$.

Finally the characteristic polynomial of a companion matrix is $p$, by direct determinant expansion. Explicitly, expansion along the first row of $tI-C$ gives the recurrencestarting from $D_1=t+a_0$. It yields $D_n=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$. ConsequentlyThis proves that a cyclic operator has equal minimal and characteristic polynomials by the cyclic basis and a direct determinant calculation, with no appeal to the Cayley-Hamilton theorem.

For $T_1$, choose the third standard basis vector $v=e_3$. Its first three orbit vectors areThey are linearly independent: their matrix of columns has determinant $-1$. They therefore span ${\mathbb{R}}^3$, proving $T_1$ has a cyclic vector, namely $e_3$.

For $T_2$, write $T_2=I+N$ where $N^2=0$. The binomial theorem gives $T_2^{j}v=v+jNv$ for every integer $j\ge 0$. The entire orbit of any $v$ lies in $\mathrm{span}\{v,Nv\}$, which has dimension of a vector space at most two. Therefore $T_2$ has no cyclic vector in the three-dimensional space. Equivalently, its minimal polynomial is $(t-1{)}^2$, while its characteristic polynomial is $(t-1{)}^3$.

Use the inner product convention that the first argument is linear and the second is conjugate-linear. This is the convention corresponding to the conjugated eigenvalue in part (c).

An orthonormal basis $(q_1,\dots ,q_n)$ is a basis with $⟨q_i,q_j⟩={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta. Each vector has norm one and distinct vectors are orthogonal.

The Gram-Schmidt process starts from a basis $(e_1,\dots ,e_n)$ and definesThe coefficients are in this order because the inner product is linear in its first argument. For $i<k$, taking the inner product with $q_i$ gives $⟨u_k,q_i⟩=0$. The residual $u_k$ cannot be zero: otherwise $e_k$ would belong to the linear span of $e_1,\dots ,e_{k-1}$, contradicting the initial basis. Normalization makes $\Vert q_k\Vert =1$, while induction preservesThus the process produces an orthonormal basis of all of $V$.

In an orthonormal basis, every vector $w$ has the coordinate expansion $w={\sum }_i⟨w,e_i⟩e_i$. If $⟨Su,v⟩=⟨u,Rv⟩$ for all $u,v$, apply it to $u=e_i$, $v=e_j$ and take complex conjugates:ConsequentlyConversely, this formula givesLinearity in the first argument and conjugate-linearity in the second extend the equality from basis vectors to all $u,v$. This proves the equivalence.

The displayed formula specifies exactly one image for each basis vector, so it defines a unique linear transformation $S^{\ast }$ and proves the existence and uniqueness of the adjoint on an inner product space.

Conjugate symmetry givesBy uniqueness of the adjoint operator, $\boxed{(S^{\ast }{)}^{\ast }=S}$. Alsoso uniqueness gives $\boxed{(ST{)}^{\ast }=T^{\ast }{S}^{\ast }}$. The reversal of order follows directly from moving the two factors across the inner product one at a time.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.

A unit in a ring is an element $u$ with a multiplicative inverse in $R$, so $uv=1$ for some $v\in R$.

An irreducible element is a nonzero nonunit $a$ such that every factorization $a=bc$ has at least one of $b,c$ a unit in a ring.

A prime element is a nonzero nonunit $p$ such that $p\mid bc$ implies $p\mid b$ or $p\mid c$. Equivalently its principal ideal $(p)$ is a nonzero prime ideal. These are definitions in the given commutative ring; the equivalence between irreducibility and primality requires additional hypotheses on the ring.

A Euclidean domain is an integral domain $R$ with a function $d:R\setminus \{0\}\to {\mathbb{N}}_0$ such that, for $a\in R$ and nonzero $b\in R$, there exist $q,r\in R$ with $a=qb+r$ and either $r=0$ or $d(r)<d(b)$. This is the Euclidean-ring convention used here.

If an ideal $I$ is zero, it is generated by zero. Otherwise choose nonzero $a\in I$ with $d(a)$ minimal. For any $b\in I$, Euclidean division gives $b=qa+r$, with $r\in I$. A nonzero remainder would have $d(r)<d(a)$, contradicting the choice. Hence $r=0$ and $b\in aR$. The reverse inclusion follows from $a\in I$, provingNow suppose $a$ is an irreducible element. Its principal ideal is proper because $a$ is not a unit. If $aR\subseteq J\subseteq R$, write $J=bR$ by the result just proved. Then $a=bc$ for some $c\in R$. Irreducibility gives either $b$ a unit, so $J=R$, or $c$ a unit, so $bR=aR$. There is no intermediate proper ideal, and therefore $aR$ is a maximal ideal.

For a field $F$, the polynomial ring $F[X]$ is an integral domain and the polynomial division algorithm gives $A=QB+R$ with $R=0$ or $\mathrm{deg}R<\mathrm{deg}B$. Thus the polynomial degree is a Euclidean function, proving $F[X]$ is a Euclidean domain.

The polynomial $X^n-2$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with prime two: all coefficients except the leading one are divisible by two, while the constant coefficient is not divisible by four. Together with Gauss's lemma, this establishes irreducibility in $\mathbb{Q}[X]$. Part (b) shows that its principal ideal is a maximal ideal, soLet $\alpha$ denote the residue class of $X$. The polynomial division algorithm gives every residue class a representative of degree less than $n$, so $1,\alpha ,\dots ,{\alpha }^{n-1}$ span $K_n$ over $\mathbb{Q}$. If a linear combination of them vanished, its degree-less-than-$n$ representative would be a multiple of the degree-$n$ polynomial $X^n-2$, and hence would be zero. They are therefore a basis, givingThe quadratic polynomial $X^2+1$ has no rational root, so it is irreducible over $\mathbb{Q}$. The same maximal-ideal argument proves $L=\mathbb{Q}[X]/(X^2+1)$ is a field.

These two quadratic fields are not isomorphic. In $L$, the class of $X$ squares to $-1$. But $K_2$ embeds in $\mathbb{R}$ by sending its class of $X$ to $\sqrt{2}$: evaluation at $\sqrt{2}$ has kernel $(X^2-2)$, by irreducibility and polynomial division. No real number squares to $-1$, so no element of $K_2$ does. A field isomorphism preserves one, minus one and squaring, so it cannot carry $L$ to $K_2$.

The Cauchy integral formula says that if $f$ is holomorphic on the simply connected open set $U$, and $\gamma$ is a positively oriented simple closed curve in $U$, then, for every $z$ in its interior,Here the contour and its enclosed region lie in $U$; the contour can be taken piecewise continuously differentiable. More generally, the same formula needs only holomorphic function on an open neighbourhood of that closed region. The orientation is counterclockwise, so the winding number about an interior point is one.

Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

The closed unit disk is a compact set, and $|f|$ is continuous on it because $f$ is holomorphic on an open neighbourhood. The extreme value theorem therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; continuity then gives the same value on the boundary. ConsequentlyThe assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.

Either equality condition says that the holomorphic function $h$ from part (i) has modulus one at an interior point: at zero in the derivative case, or at the given nonzero point in the other case. Since $|h|\le 1$ everywhere, the maximum modulus principle makes $h$ a constant $w$ with $|w|=1$. ThereforeThese maps are rotations of the unit disk and do achieve both equalities, so this also identifies all equality cases of the Schwarz lemma.

The Laurent theorem states that a holomorphic function on the annulus $R<|z-a|<S$ has a unique Laurent seriesThe integration circle is counterclockwise. The series converges absolutely at every point of the annulus and uniformly on every compact subannulus. The coefficient integrals are independent of the choice of $\rho$ by the Cauchy integral theorem applied between two such circles. These statements include $R=0$ or $S=\infty$ when the indicated annulus has those endpoints.

An isolated singularity at $z_0$ means that some radius $r>0$ has $f$ holomorphic on the punctured disk $0<|z-z_0|<r$, while $f$ is not initially defined as a holomorphic function at $z_0$. The adjective isolated refers to the absence of other singularities in this punctured neighbourhood; it permits a removable singularity.

The point is a removable singularity if assigning a value to $f(z_0)$ makes $f$ holomorphic on a full neighbourhood of $z_0$. Equivalently, its Laurent series has no nonzero coefficients of negative powers. The assigned value is its constant coefficient, or equivalently the finite limit ${\lim }_{z\to z_0}f(z)$.

A pole of order $n\ge 1$ meansEquivalently, the Laurent series has a finite nonzero principal part, with lowest power exactly $(z-z_0{)}^{-n}$. Thus $(z-z_0{)}^{n}f(z)$ extends holomorphically with nonzero value at $z_0$; using a smaller positive power would not remove the singularity.

An essential singularity is an isolated singularity that is neither a removable singularity nor a pole. In its Laurent series, infinitely many coefficients of negative powers are nonzero. The alternatives follow because no negative terms give a removable singularity, while finitely many with a lowest nonzero power give a pole.

Inside the unit circle, use the geometric series for $1/(1-z)$:Outside the unit circle, expand in powers of $1/z$ instead:The different Laurent series reflect different annuli separated by the pole at one. Both follow from absolutely convergent geometric series in their stated regions.

Take the Laurent series of $f$ on $0<|z|<r$. For every integer $k\ge 1$, its coefficient of $z^{-k}$ satisfies, on a circle of radius $0<\rho <r$,The coefficient is independent of $\rho$. Letting $\rho \downarrow 0$ therefore gives $c_{-k}=0$ for every $k\ge 1$. The Laurent series is a power series with no negative powers and supplies a holomorphic extension at zero. Thus zero is a removable singularity. This is the Riemann removable singularity theorem, with boundedness used to eliminate the entire principal part.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.

The tangent half-angle substitution $u=\tan (t/2)$ maps $0\le t<\pi$ to $0\le u<\infty$, withBecause $a>1$, both $a+1$ and $a-1$ are positive. ThusEvaluating the improper integral givesAlthough the change of variable has an infinite endpoint, the original integrand is continuous on the closed interval, and the transformed integrand decays as $u^{-2}$, so the limiting substitution is justified.

Use the branch of the complex logarithm with $0<\arg z<2\pi$, so the branch cut is the positive real axis, and take $\sqrt{z}=\exp (\mathrm{Log}z/2)$ on the same slit plane. Apply the residue theorem toon a positively oriented keyhole contour with inner radius $\rho <1$ and outer radius $R>1$. The upper bank travels from $\rho$ to $R$, the large circle counterclockwise, the lower bank from $R$ to $\rho$, and the small circle clockwise. One may first keep the banks at small nonzero angles and then take their limits.

On the upper bank, $\mathrm{Log}x=\log x$ and $\sqrt{x}$ is positive. On the lower bank, the corresponding limits are $\log x+2\pi i$ and $-\sqrt{x}$. Reversing the lower-bank orientation therefore makes the sum of the bank integrals tend toBoth are absolutely convergent: near zero compare with $x^{-1/2}(1+|\log x|)$, and near infinity with $x^{-5/2}(1+\log x)$. On the large circle, $|z^2+1|\ge R^2-1$, so its contribution is $O((\log R+1)R^{-3/2})$. On the small circle it is $O(\sqrt{\rho }(1+|\log \rho |))$. Both tend to zero.

The two simple poles are $i$ and $-i$. On the chosen branch their arguments are $\pi /2$ and $3\pi /2$, respectively. Consequently their residues areFor example, the first follows from $\mathrm{Log}i=i\pi /2$, $\sqrt{i}=e^{i\pi /4}$ and the derivative $2i$ of the denominator factor $z^2+1$. The second uses $\sqrt{-i}=e^{3\pi i/4}$, rather than the value from the principal square-root branch. Adding and applying the residue theorem yieldsAs $J$ is real, taking real parts givesThe imaginary part also gives $J=\pi /\sqrt{2}$, an independent consistency check on the orientation and branch factors.

A critical point of a dynamical system or equilibrium is a point $(a,b)$ with $f(a,b)=g(a,b)=0$, so the constant function $(x,y)=(a,b)$ is a solution. Put $u=x-a$, $v=y-b$. For a continuously differentiable vector field, its linearization iswith remainder $o(\sqrt{u^2+v^2})$ in the original nonlinear system. For distinct eigenvalues, $M$ is diagonalizable over $\mathbb{C}$, and the linear solutions are combinations of $e^{{\lambda }_{j}t}{w}_j$, where $w_j$ are corresponding eigenvectors. Real parts determine growth or decay; nonzero imaginary parts give rotation. For hyperbolic equilibria, meaning no eigenvalue has zero real part, the Hartman-Grobman theorem transfers the local stability and qualitative orbit pattern to the nonlinear system. A purely imaginary pair is not hyperbolic: it determines a linear centre, while higher-order terms decide whether the nonlinear system actually has a centre.

A stable node hasBoth exponential modes decay as $t\to \infty$, so the equilibrium is asymptotically stable. Generic linear orbits approach tangent to the eigenvector for the less negative eigenvalue; the eigenvector directions themselves are invariant straight lines. There is no rotation because both eigenvalues are real.

An unstable node hasBoth linear modes grow forward in time. Nearby nonconstant orbits move away from the equilibrium, and approach it under reversed time; their local directions come from the two real eigenvectors.

A saddle equilibrium hasThe negative eigenvalue gives a stable manifold direction and the positive eigenvalue an unstable manifold direction. Perturbations along the latter grow, so the equilibrium is unstable even though there are special trajectories approaching it.

A centre equilibrium of the linearized real planar system hasAfter a real invertible linear change of coordinates, the system is a constant angular rotation. Its nonzero orbits are ellipses of period $2\pi /|\omega |$, so it is stable but not asymptotically stable. This is a classification of the linearization: a centre in the linearization need not be a nonlinear centre. In particular, imaginary eigenvalues alone do not prove a nonlinear centre equilibrium, which requires a punctured neighbourhood filled with periodic orbits. Part (c) illustrates exactly this limitation.

A stable spiral has the complex conjugate eigenvaluesIn suitable real coordinates the radius decays like $e^{\alpha t}$ while the angle changes at rate $\beta$. The orbit therefore spirals towards the equilibrium, which is asymptotically stable. This is also called a stable focus.

An unstable spiral hasRotation is accompanied by exponentially increasing radius in suitable real coordinates. Thus nearby nonconstant orbits spiral away, and the equilibrium is unstable. This is also called an unstable focus.

The first component of the vector field vanishes only when $y=0$ or $x=\pm 1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five equilibria. Their Jacobian matrix isAt $(3,0)$,The real parts are negative and the imaginary parts are nonzero, so $(3,0)$ is a stable spiral.

At $(1,1)$ the Jacobian matrix is $\mathrm{diag}(2,-1)$. Its real eigenvalues have opposite signs, so $(1,1)$ is a saddle equilibrium.

At $(1,2)$,Both eigenvalues are positive, so $(1,2)$ is an unstable node.

At $(-1,1)$ the Jacobian matrix is $\mathrm{diag}(-2,-3)$, giving a stable node.

At $(-1,4)$,so $(-1,4)$ is a saddle equilibrium. All five equilibria are hyperbolic, so these linear classifications determine their nonlinear local stability as well.

The Bendixson-Dulac theorem states that if $(P,Q)$ is a continuously differentiable planar vector field on a simply connected open region $D$, and a continuously differentiable Dulac function $B$ makeshave one weak sign throughout $D$, without vanishing identically on any nonempty open subset, then there is no nonconstant periodic orbit contained in $D$. To see why, a hypothetical simple periodic orbit bounds a region in $D$. The divergence theorem gives the integral of this expression as the outward flux of $B(P,Q)$ across the orbit. The flux is zero because the vector field is tangent to its orbit, whereas the one-sign area integral is nonzero, a contradiction. The usual stricter version assumes a strictly positive or strictly negative expression everywhere.

For the printed polynomial system, the origin is an equilibrium andThus its linearization is a centre equilibrium. For the full system take $B=1$. The divergence isOn the open disk $x^2+y^2<3$ this is nonpositive, and is negative wherever $x\ne 0$. Its zero set inside the disk is a line, so it cannot vanish on any open subset. The Bendixson-Dulac theorem therefore provesThis also exposes a terminology problem in the source: a nonlinear centre equilibrium has surrounding closed orbits, so the printed claim of a centre cannot hold in that sense. The imaginary eigenvalues establish only a centre of the linearized system. In fact the full equilibrium is an asymptotically stable weak focus, as the following direct check shows.

A quartic Lyapunov correction for a weak focus makes the nonlinear damping visible. Put $r^2=x^2+y^2$ and consider the Lyapunov functionIt is positive definite sufficiently close to zero. Differentiating along the polynomial vector field givesFor clarity, the quartic terms come from $-x^4$ in the derivative of $r^2/2$, and from applying $y{\partial }_x-x{\partial }_y$ to the quartic correction. Their sum is $-3x^4/8-3x^2{y}^2/4-3y^4/8$; all other terms have degree at least six. Therefore $\dot{V}<0$ in a sufficiently small punctured neighbourhood, proving asymptotic stability by Second Lyapunov theorem. Finallyso nonconstant nearby orbits turn clockwise as they converge. The equilibrium is a weak focus despite the centre in its linearization. This interpretation resolves both the local dynamics and the absence of closed orbits without changing the printed coefficients.

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.